Class 11

### Find the vector and Cartesian equations of a plane which is at a distance of 7 units from the origin and whose normal vector from the origin is

Answer: Given, $\begin{array}{l} d = 7\\ \overline n = 3\widehat i + 5\widehat j - 6\widehat k \end{array}$ The unit vector normal to the plane:              ...

### Find the distance of the point (2, 1, 0) from the plane 2x + y – 2z + 5 = 0.

Answer: Given plane, 2x + y – 2z + 5 = 0 The point is (2, 1, 0).

### Find the distance of the point (1, 1, 2) from the plane plane .

Answer: Given plane, $\overrightarrow r .(2\widehat i - 2\widehat j + 4\widehat k) + 5 = 0$ The cartesian form: 2x – 2y + 4z + 5 = 0 The point is (1, 1, 2).

### Find the distance of the point (3, 4, 5) from the plane .

Answer: Given plane, $\overrightarrow r .(2\widehat i - 5\widehat j + 3\widehat k) = 13$ The cartesian form: 2x – 5y + 3z – 13 = 0 The point is (3, 4, 5).

### Find the distance of the point from the plane .

Answer: Given plane, $\overrightarrow r .(\widehat i + \widehat j + \widehat k) + 17 = 0$ The cartesian form: x + y + z +17 = 0 The point is $(\widehat i + 2\widehat j + 5\widehat k)$ => (1, 2,...

### Find the distance of the point from the plane .

Answer: Given plane, $\overrightarrow r .(3\widehat i - 4\widehat j + 12\widehat k) = 9$ The cartesian form: 3x – 4y + 12z – 9 =0 The point is $(2\widehat i - \widehat j - 4\widehat k)$ => (2,...

### Find the vector and Cartesian equations of a plane which passes through the point (1, 4, 6) and normal vector to the plane is is

Answer: Given, A = (1, 4, 6)

### Find the vector and Cartesian equations of a plane which is at a distance of 6/√29 from the origin and whose normal vector from the origin is is

Answer: Given,                                 = (x × 2) + (y × (-3)) + (z × 4) = 2x - 3y + 4z The Cartesian equation...

### Find the vector equation of a plane which is at a distance of 5 units from the origin and which has as the unit vector normal to it.

Answer: Given, $\begin{array}{l} d = 5\\ \widehat n = \widehat k \end{array}$ The equation of plane at 5 units distance from the origin $\begin{array}{l} \widehat n \end{array}$ and as a unit...

### Reduce the equation of the plane 4x – 3y + 2z = 12 to the intercept form, and hence find the intercepts made by the plane with the coordinate axes.

Answer: Equation of the plane: 4x – 3y + 2z = 12 $\begin{array}{l} \frac{4}{{12}}x - \frac{3}{{12}}y + \frac{2}{{12}}z = 1\\ \frac{x}{3} + \frac{y}{{ - 4}} + \frac{z}{6} = 1 \end{array}$ It is the...

### Write the equation of the plane whose intercepts on the coordinate axes are 2, – 4 and 5 respectively.

Answer: Given, Coordinate axes are 2, - 4, 5 The equation of the variable plane:         The required equation of the plane is 10x – 5y + 4z = 20.

### Find the equation of the plane passing through each group of points: A (-2, 6, -6), B (-3, 10, -9) and C (-5, 0, -6)

Answer: Given, A (-2, 6, -6) B (-3, 10, -9) C (-5, 0, -6)                                        ...

### 3. Show that the four points A (0, -1, 0), B (2, 1, -1), C (1, 1, 1) and D (3, 3, 0) are coplanar. Find the equation of the plane containing them.

Answer: Given, A (0, -1, 0) B (2, 1, -1) C (1, 1, 1) D (3, 3, 0)                   4x – 3 (y + 1) + 2z = 0 4x – 3y + 2z – 3 = 0 Take x = 0, y = 3 and z =...

### Show that the four points A (3, 2, -5), B (-1, 4, -3), C (-3, 8, -5) and D (-3, 2, 1) are coplanar. Find the equation of the plane containing them.

Answer: Let us take, The equation of the plane passing through A (3, 2, -5) a (x – 3) + b (y – 2) + c (z + 5) = 0 It passes through the points B (-1, \4, -3) and C (-3, 8, -5) a (1 – 3) + b (4 – 2)...

### Find the equation of the plane passing through each group of points: (i) A (2, 2, -1), B (3, 4, 2) and C (7, 0, 6) (ii) A (0, -1, -1), B (4, 5, 1) and C (3, 9, 4)

Answer: (i) Given, A (2, 2, -1) B (3, 4, 2) C (7, 0, 6)                                        ...

### Using properties of determinants prove that:

Solution: $\left|\begin{array}{ccc}a^{2} & b^{2} & c^{2} \\ (a+1)^{2} & (b+1)^{2} & (c+1)^{2} \\ (a-1)^{2} & (b-1)^{2} & (c-1)^{2}\end{array}\right|$...

### Prove that

Answer: = cosx cos2x cos4x cos8x Multiply and divide by 2sinx, we get We know that, sin 2x = 2 sinx cosx Replacing x by 2x, we get sin 2(2x) = 2 sin(2x) cos(2x) or sin 4x = 2 sin 2x cos 2x...

### Prove that

Answer: Taking sinx common from the numerator and cosx from the denominator

### Prove that:

Answer: = RHS ∴ LHS = RHS Hence Proved

### Prove that:

Answer: Multiply and divide by 2, we get = RHS ∴ LHS = RHS Hence Proved

### Prove that cot x – 2cot 2x = tan x

Answer: Taking LHS, = cot x – 2cot 2x …(i) We know that, cot x = cos x/ sin x Replacing x by 2x, we get cot 2x = cos 2x/ sin 2x So, eq. (i) becomes = sin x/ cos x = tan x = RHS ∴ LHS = RHS Hence...

### Prove that cos 2x + 2sin2 x = 1

Answer: Taking LHS = 2 – 1 = 1 = RHS

### Prove that cosec 2x + cot 2x = cot x

Answer: To Prove: cosec 2x + cot 2x = cot x Taking LHS, = cosec 2x + cot 2x …(i) We know that, cosec x = 1 / sin x and cot x = cos x/ sin x Replacing x by 2x, we get = cos x/ sinx = cot...

### Prove that sin 2x(tan x + cot x) = 2

Answer: Taking LHS, sin 2x(tan x + cot x) We know that: We know that, sin 2x = 2 sinx cosx = 2 = RHS ∴ LHS = RHS Hence Proved

### Prove that:

Answer: = sin x / cos x = tan x = RHS ∴ LHS = RHS Hence Proved

### Find the area of a parallelogram with base equal to 25 cm and the corresponding height measuring 16.8 cm.

Sol: Given: Base = 25 cm Height = 16.8 cm \Area of the parallelogram = Base ´ Height = 25cm ´16.8 cm = 420 cm2

### In a bulb factory, machines A, B and C manufactures and bulbs respectively. Out of these bulbs and of the bulbs produced respectively by and are found to be defective. A bulb is picked up at random from the total production and found to be defective. Find the probability that this bulb was produced by machine .

Let $A$ : Manufactured from machine $A$, B : Manufactured from machine B C: Manufactured from machine C D : Defective bulb We want to find $P(A \mid D)$, i.e. probability of selected defective bulb...

### There are three boxes, the first one containing 1 white, 2 red and 3 black balls; the second one containing 2 white, 3 red and 1 black ball and the third one containing 3 white, 1 red and 2 black balls. A box is chosen at random, and from it, two balls are drawn at random. One ball is red and the other, white. What is the probability that they come from the second box?

let $A:$ Ball drawn from bag $A$ B : Ball is drawn from bag B $C:$ Ball is drawn from bag $C$ BB: Black ball WB: White ball RB: Red ball Assuming, selecting bags is of equal probability i.e....

### Mark the tick against the correct answer in the following: The principal value of is A. B. C. D. none of these

Solution: Option(A) is correct. To Find: The Principle value of $\operatorname{cosec}^{-1}(-\sqrt{2})$ Let the principle value be given by $\mathrm{x}$ Now, let...

### Find the equation of the ellipse with eccentricity     , foci on the y-axis, center at the origin and passing through the point     .

Given Eccentricity = $\frac{3}{4}$ We know that Eccentricity = c/a Therefore,c=$\frac{3}{4}$a

### Find the equation of the ellipse with center at the origin, the major axis on the x-axis and passing through the points     and     .

Given: Center is at the origin and Major axis is along x – axis So, Equation of ellipse is of the form $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$…(i) Given that ellipse passing...

### Find the equation of the ellipse whose foci are at     and

Given: Coordinates of foci = $\left( \mathbf{0},\text{ }\pm \mathbf{4} \right)$ …(i) We know that, Coordinates of foci = $\left( 0,\text{ }\pm c \right)$ …(ii) The coordinates of the foci are...

### Find the equation of the ellipse whose foci are at     and e=1/2

Let the equation of the required ellipse be $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$ Given: Coordinates of foci = $\left( \pm 1,\text{ }0 \right)$ …(i) We know that,...

### Find the equation of the ellipse whose foci are     and the eccentricity is

Let the equation of the required ellipse be Given: Coordinates of foci = $\left( \pm 2,\text{ }0 \right)$…(iii) We know that, Coordinates of foci = $\left( \pm c,\text{ }0 \right)$…(iv) ∴ From...

### Find the equation of the ellipse the ends of whose major and minor axes are     and     respectively.

Given: Ends of Major Axis = $\left( \pm \mathbf{4},\text{ }\mathbf{0} \right)$ and Ends of Minor Axis = $\left( \mathbf{0},\text{ }\pm \mathbf{3} \right)$ Here, we can see that the major axis is...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: $\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}$ Divide by $100$ to both the sides, we get...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: $\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}$ Divide by $100$ to both the sides, we get...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: $\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}$ Divide by $100$ to both the sides, we get...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: $\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}$ Divide by $16$ to both the sides, we get...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: $\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}$ Divide by $16$ to both the sides, we get...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: $\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}$ Divide by $16$ to both the sides, we get...

### Prove that:

Answer: = cot x = RHS ∴ LHS = RHS Hence Proved

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: $\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{36}$ Divide by $36$ to both the sides, we get $\frac{9}{36}{{x}^{2}}+\frac{1}{36}{{y}^{2}}=1$...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: $\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{36}$ Divide by $36$ to both the sides, we get $\frac{9}{36}{{x}^{2}}+\frac{1}{36}{{y}^{2}}=1$...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: $\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{36}$ Divide by $36$ to both the sides, we get $\frac{9}{36}{{x}^{2}}+\frac{1}{36}{{y}^{2}}=1$...

### Prove that:

Answer: = tan x = RHS ∴ LHS = RHS Hence Proved

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: $\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{2}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{18}$…(i) Divide by $18$ to both the sides, we get...

### Prove that:

Answer: Taking LHS = cos x + sin x = RHS ∴ LHS = RHS Hence Proved

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: $\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{2}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{18}$…(i) Divide by $18$ to both the sides, we get...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: $\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{2}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{18}$…(i) Divide by $18$ to both the sides, we get...

### If cos x = -1/3 , find the value of cos 3x

Answer; We know that, cos 3x = 4cos3 x – 3 cosx Putting the values, we get

### If sinx = 1/6, find the value of sin 3x.

Answer: To find: sin 3x We know that, sin 3x = 3 sinx – sin3 x Putting the values, we get

### Find the (v) length of the latus rectum of each of the following ellipses.     Answer :

Given:  $\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1$….(i) Since, $9<16$ So, above equation is of the form, $\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1$…(ii) Comparing eq. (i)...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given:  $\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1$….(i) Since, $9<16$ So, above equation is of the form, $\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1$…(ii) Comparing...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given:  $\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1$….(i) Since, $9<16$ So, above equation is of the form, $\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1$…(ii) Comparing eq. (i)...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: $\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1$…(i) Since, $4\text{ }<\text{ }25$ So, above equation is of the form, $\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1$…(ii)...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: $\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1$…(i) Since, $4\text{ }<\text{ }25$ So, above equation is of the form, $\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1$…(ii)...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: $\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1$…(i) Since, $4\text{ }<\text{ }25$ So, above equation is of the form, $\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1$…(ii)...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: $\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}$ $\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1$….(i) Since,...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: $\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}$ $\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1$….(i) Since,...